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Re: [Phys-L] consistency: 1/r^2 electrostatics, 1/r radiation field



Go into an "oscillating frame" and look at the field of a charge which is stationary in some inertial frame.
For JD this should be a simple transformation :>)))

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
treborsci@verizon.net
www.sciamanda.com

Hi Folks --

Here's something fun to think about. Consider the contrast:

a) In electromagnetic radiation, the E-field of a point source
dies off like 1/r. The power and the energy density scale like
E squared, so they scale like 1/r^2, which is consistent with the
idea that energy is conserved as the wave spreads out.

b) In electrostatics, the E-field of a point source dies off
like 1/r^2. The energy density falls off like 1/r^4, but it isn’t
/transporting/ any energy, so conservation doesn’t have anything
to say about it.

The question arises, can we obtain a consistent view of these two
facts? This is not going to be easy, because starting with the
1/r^2 Coulomb field of a point charge, I don’t see any way to
explain the 1/r radiation field. By way of contrast, if there is
some sort of cancellation, I can arrange something that falls off
/faster/ than 1/r^2 – such as a dipole field that falls off like
1/r^3 – but I cannot cook up anything that falls of slower than
1/r^2. I’ve seen a number of books that claim to explain things
this way, but it never made any sense to me.

So some profound questions remain:

a) Can we take the low-frequency limit of the radiation field
and recover the Coulomb field?
b) Can we wiggle the Coulomb field and get the radiation field?>c) Or are they related in some other way?
The short answers are: (a) no, (b) no, and (c) yes



The answers (a) and (b) become obvious from a very simple observation: The Coulomb field is radial (pointing away from or to the source), whereas the radiation field is perpendicular to its propagation direction which is away from the source. The two fields differ not only quantitatively, but also on the qualitative level, so there is no continuous transition between them. As to part (c), - both kinds of field are parts of general solution of EM field equations.
Moses Fayngold,NJIT


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